If two waves represented by y1-4 sin- t and y2-3 sin - t-3 interfere at a point- the amplitude of the resulting wave will be close to

Given, y_1=4sinω t y_2= 3sin (ω t+π3) Here, phase difference, ϕ =π3 A_1=4, A_2=3 So, Resultant Amplitude (A_R) is given by, A_R=√(A_1^2+A_2^2+2A_1A_2cosϕ) ...

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Standard XII

Physics

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If two waves ...

Question

If two waves represented by $y_{1} = 4 sin ω t$ and $y_{2} = 3 sin (ω t + \frac{π}{3})$ interfere at a point, the amplitude of the resulting wave will be close to

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Solution

The correct option is B $6$
Given,
$y_{1}$ = $4 sin ω t$
$y_{2}$ = $3 sin (ω t + \frac{π}{3})$

Here, phase difference, $ϕ = \frac{π}{3}$

$A_{1} = 4$ , $A_{2} = 3$

So, Resultant Amplitude $(A_{R})$ is given by,

$A_{R} = \sqrt{A_{1}^{2} + A_{2}^{2} + 2 A_{1} A_{2} cos ϕ}$

$A_{R} = \sqrt{(4)^{2} + (3)^{2} + (2 \times 4 \times 3 \times cos \frac{π}{3}})$

$A_{R} = \sqrt{37} \approx 6$

Hence, option $(B)$ is correct.

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If two waves represented by y1=4sinωt and y2=3sin(ωt+π3) interfere at a point, the amplitude of the resulting wave will be close to

The correct option is B 6Given, y1=4sinωt y2= 3sin(ωt+π3) Here, phase difference, ϕ=π3 A1=4, A2=3 So, Resultant Amplitude (AR) is given by, AR=√A21+A22+2A1A2cosϕ AR=√(4)2+(3)2+(2×4×3×cosπ3) AR=√37≈6 Hence, option (B) is correct.

If two waves represented by $y_{1} = 4 sin ω t$ and $y_{2} = 3 sin (ω t + \frac{π}{3})$ interfere at a point, the amplitude of the resulting wave will be close to